This is another exercise from Allan's book "Introduction to Banach Spaces and Algebras".

Exercise 2.9: A Banach space $E$ is said to be homeomorphic to its square if $E\oplus E$ is linearly homeomorphic to $E$. Prove that the spaces $c$ and $C([0,1])$ have this property.

Note that we haven't developed much theory yet (in this book!) Well, $c_0 \oplus c_0 \cong c_0$ (split a sequence into the parts supported on the evens and odds say) and $c\cong c_0$, so that does the case of $c$. I cannot see an elementary proof for $C([0,1])$. Here's a less than easy proof: clearly $C([0,1]) \oplus_\infty C([0,1]) = C([0,1] \cup [2,3])$ say. Then appeal to Miljutin's Theorem, as $[0,1] \cup [2,3]$ is an uncountable compact metric space, $C([0,1] \cup [2,3]) \cong C([0,1])$.

Surely I don't need the full power of Miljutin's Theorem. But, for example, $C([0,1] \cup [2,3])$ cannot be isometric to $C([0,1])$ (if we use real scalars) by the Banach-Stone theorem, as $[0,1] \cup [2,3]$ is not homeomorphic to $[0,1]$.

Surely using Miljutin's Theorem is not what the book intends.

Is there an elementary proof that $C([0,1]) \oplus C([0,1]) \cong C([0,1])$?

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    $\begingroup$ Perhaps this is a start: Let $T:C([0,1])\oplus C([0,1])\to C([0,1])\oplus \mathbb R$ be defined by $T(f,g)=(h(t),g(0))$ where $$h(t)=\begin{cases} f(2t) &\mbox{if } t \leq 1/2 \\ g(2t-1)+f(1)-g(0) & \mbox{if } t>1/2 \end{cases} $$ and note that $h$ is continuous as $h(1/2)=f(1)=g(1)+f(1)-g(1)$. $\endgroup$ – Alex Becker Mar 5 '12 at 11:38
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    $\begingroup$ This was known to Banach, and is in his book; see Theorem 6 and Theorem 8 on p.111 on the English language translation of the book. $\endgroup$ – Philip Brooker Mar 5 '12 at 12:39
  • $\begingroup$ @PhilipBrooker Could you give a reference? $\endgroup$ – Norbert Mar 5 '12 at 13:02
  • $\begingroup$ @Phil: Indeed! That certainly counts as "elementary". But it does seem a stretch that someone could think of this on their own... So now I do wonder if the exercise was set erroneously. If you want to make that into an answer, I'll accept it... $\endgroup$ – Matthew Daws Mar 5 '12 at 13:12
  • $\begingroup$ @Matt: Does it really say "homeomorphic to it square" rather than "isomorphic to its square" in Allan's book? I only just noticed that wording in your question, and it is obviously strange because all separable Banach spaces are homeomorphic to one another... $\endgroup$ – Philip Brooker Mar 6 '12 at 9:33

(Reposting my comment from above:) This result was known to Banach and a proof can be found in his book Theory of Linear Operations (note that this is the English translation of the French original); for details see Theorems 6 and 8 on p.111.

To give a little more information here, Theorem 6 from Banach's book asserts that $C([0,1])$ is isomorphic to $C([0,1])\oplus c_0$, whilst Theorem 8 from the same book shows similar to Alex Becker's comment above that $C([0,1])\oplus C([0,1])$ is isomorphic to $C([0,1])\oplus \mathbb{R}$. One then deduces that $$C([0,1])\oplus C([0,1]) \approx C([0,1])\oplus \mathbb{R}\approx (C([0,1])\oplus c_0) \oplus \mathbb{R} $$ $$\approx C([0,1])\oplus (c_0 \oplus \mathbb{R}) \approx C([0,1])\oplus c_0 \approx C([0,1]).$$

As Matt points out in the comments above, coming up with and executing this strategy seems a little heavy for an exercise!

  • $\begingroup$ Thanks! That'll nicely move your reputation up from "666"... ;-> $\endgroup$ – Matthew Daws Mar 5 '12 at 14:52
  • $\begingroup$ Haha! Thanks Matt. I answered a question yesterday but even then got no votes for my answer, so I was stuck on that particular score for a while :-) . Thanks again for 'saving' me! $\endgroup$ – Philip Brooker Mar 5 '12 at 14:57

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